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Let $q$ be prime and $k$ be a natural number. When does $\sigma(q^k)$ have a prime factor greater than $q$?

We can slightly reduce the problem by noting that $$\sigma(q^k)=\frac{q^{k+1}-1}{q-1}$$ Thus if $p\nmid q-1$, then $p\mid \sigma(q^k)$ if and only if $p\mid q^{k+1}-1$. Clearly a prime $p>q$ does not divide $q-1$, so $\sigma(q^k)$ has a prime factor greater than $q$ if and only if $q^{k+1}-1$ does.

Using this restatement, if one takes a look at one of my answers to this question we see this is the case when $k+1\ge q$ letting $a=q$, $b=1$, and $n=k+1$ (note the question requires $b>1$, but the proof only uses Zsigmondy's Theorem in which $b>0$ suffices). However clearly this is true in more than just this case. Can we describe all cases in which this is true, or at least expand the set of $(q,k)$ for which we know that this holds?

Edit
Lets define $\kappa(n)$ to be the smallest prime congruent to $1\mod n$. A modified argument of that in the linked question, where we restrict $b$ to $1$, leads one to conclude that there exists a prime $\ge \kappa(n)$ that divides $q^n-1$. Thus if $$\kappa(k+1)\ge q$$ (or really $\kappa(k+1)\ge q-1$ since $q\nmid q^n-1$), then $\sigma(q^k)$ satisfies the problem, slightly increasing the set of valid $(q,k)$.

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  • $\begingroup$ When $k+1$ is even, Because number of terms of series representing sum of divisors is equal to $k+1$ and if it is even then there is a common factor like $q+1$ in it; $\sigma=[q^k(q+1)+q^{k-1}(q+1)+ . . . +(q+1) ]$. $\endgroup$ – sirous Feb 18 at 12:21
  • $\begingroup$ @sirous However, unless $q=2$, since $q$ is a prime $2\mid q+1$ and so any prime factors contributed by $q+1$ will be smaller than $q$. $\endgroup$ – Will Fisher Feb 18 at 17:42
  • $\begingroup$ the only even prime i.e. 2 always is disturbing. What I claim is for odd primes and is correct. $\endgroup$ – sirous Feb 19 at 6:31
  • $\begingroup$ Wont you down vote me if I post this as an answer? $\endgroup$ – sirous Feb 19 at 11:53
  • $\begingroup$ @sirous the thing is you found a factor but I’m looking for prime factors. $\endgroup$ – Will Fisher Feb 19 at 12:53

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